Упр.1134 ГДЗ Никольский Потапов 6 класс (Математика)

Задача

а) 11/15•(4 1/2-3 2/5 :17/20)+1 11/20;
б) 5 4/7 :1 5/21-(5 2/15•3/22+1 14/15);
в) 7 2/3+4 1/6•(6 2/7-3 5/7);
г) 4 2/7 :1 5/21+(4 3/13•14/15-3 1/3).

Подробный ответ

а) Преобразуем смешанные числа в неправильные дроби и выполняем действия по порядку:

$$ \frac{11}{15}\cdot\left(4\frac{1}{2}-3\frac{2}{5}:\frac{17}{20}\right)+1\frac{11}{20} = \frac{11}{15}\cdot\left(\frac{9}{2}-\frac{17}{5}\cdot\frac{20}{17}\right)+\frac{31}{20} $$
$$ = \frac{11}{15}\cdot\left(\frac{9}{2}-4\right)+\frac{31}{20} = \frac{11}{15}\cdot\frac{1}{2}+\frac{31}{20} = \frac{11}{30}+\frac{31}{20} = \frac{22}{60}+\frac{93}{60} = \frac{115}{60} = 1\frac{11}{12} $$

б)

$$ 5\frac{4}{7}:1\frac{5}{21}-\left(5\frac{2}{15}\cdot\frac{3}{22}+1\frac{14}{15}\right) = \frac{39}{7}:\frac{26}{21}-\left(\frac{77}{15}\cdot\frac{3}{22}+\frac{29}{15}\right) $$
$$ = \frac{39}{7}\cdot\frac{21}{26}-\left(\frac{7}{5}\cdot\frac{1}{2}+\frac{29}{15}\right) = \frac{3}{1}\cdot\frac{3}{2}-\left(\frac{7}{10}+\frac{29}{15}\right) $$
$$ = \frac{9}{2}-\left(\frac{21}{30}+\frac{58}{30}\right) = \frac{9}{2}-\frac{79}{30} = \frac{135}{30}-\frac{79}{30} = \frac{56}{30} = 1\frac{13}{15} $$

в)

$$ 7\frac{2}{3}+4\frac{1}{6}\cdot\left(6\frac{2}{7}-3\frac{5}{7}\right) = 7\frac{2}{3}+\frac{25}{6}\cdot\left(\frac{44}{7}-\frac{26}{7}\right) $$
$$ = 7\frac{2}{3}+\frac{25}{6}\cdot\frac{18}{7} = 7\frac{2}{3}+\frac{75}{7} = 7\frac{2}{3}+10\frac{5}{7} = 17\frac{29}{21} = 18\frac{8}{21} $$

г)

$$ 4\frac{2}{7}:1\frac{5}{21}+\left(4\frac{3}{13}\cdot\frac{14}{15}-3\frac{1}{3}\right) = \frac{30}{7}:\frac{26}{21}+\left(\frac{55}{13}\cdot\frac{14}{15}-\frac{10}{3}\right) $$
$$ = \frac{30}{7}\cdot\frac{21}{26}+\left(\frac{11}{13}\cdot\frac{14}{3}-\frac{10}{3}\right) = \frac{45}{13}+\left(\frac{154}{39}-\frac{130}{39}\right) $$
$$ = \frac{45}{13}+\frac{24}{39} = 3\frac{6}{13}+\frac{24}{39} = 3\frac{18}{39}+\frac{24}{39} = 3\frac{42}{39} = 4\frac{3}{13} $$